An ideal gas receives 425 J of heat and expands by 1.17 L against an external pressure of 100.0 kPa. What is the change in internal energy of the system?

Use-full formula...!

Δ
W = p
Δ
V

Question

An ideal gas receives 425 J of heat and expands by 1.17 L against an external pressure of 100.0 kPa.  What is the change in internal energy of the system?

Use-full formula...!

Δ
W = p 
Δ
V

BlackTom AI · Accepted Answer

Using the first law of thermodynamics: ΔU = q − W, where q is heat added to the system and W is the work done by the system on the surroundings. The problem gives q = +425 J (heat received) and asks for W = pΔV.

Calculate the work:
- Given p = 100.0 kPa and ΔV = 1.17 L.
- Use either unit-conversion approach:
  Method A (kPa·L → J): 1 kPa·L = 1 J, so W = 100.0 kPa × 1.17 L = 117.0 kPa·L = 117.0 J.
  Method B (SI base units): p = 100.0 kPa = 100000 Pa, ΔV = 1.17 L = 1.17 × 10^−3 m^3, so W = pΔV = 100000 Pa × 1.17×10^−3 m^3 = 117.0 J.
Either method yields W = 117.0 J done by the gas.

Apply the first law:
ΔU = q − W = 425.0 J − 117.0 J = 308.0 J.

Why this sign convention? Because the system receives 425 J of heat (increasing internal energy) but does 117 J of work on the surroundings (energy leaving the system). The net internal-energy increase is the difference, +308 J.

Common errors to avoid:
- Forgetting to convert liters to cubic meters (or using kPa·L correctly) leads to unit errors by a factor of 1000.
- Using the wrong sign for work (some conventions define W as work done on the system; here W is work done by the system, so it is subtracted).
- Rounding too early; keep consistent significant figures.

Conclusion: The change in internal energy is an increase of 308 J, which matches the correct answer.

An ideal gas receives 425 J of heat and expands by 1.17 L against an external pressure of 100.0 kPa. What is the change in internal energy of the system? Use-full formula...! Δ W = p Δ V简答题

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